$f(x,y) = 5x^4 - 2y^2$ What is $\dfrac{\partial f}{\partial y}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $20x^3$ (Choice C) C $-4y^2$ (Choice D) D $-4y$
We want to find $\dfrac{\partial f}{\partial y}$, which is the partial derivative of $f$ with respect to $y$. When we take a partial derivative with respect to $y$, we treat $x$ as if it were a constant. Let's break $f(x, y)$ down term by term. $\begin{aligned} &\dfrac{\partial}{\partial y} \left[ 5x^4 \right] = 0 \\ \\ &\dfrac{\partial}{\partial y} \left[ -2y^2 \right] = -4y \end{aligned}$ Adding the terms back together, we get the partial derivative. In conclusion: $\dfrac{\partial f}{\partial y} = 0 - 4y = -4y$